# A projectile is shot from the ground at an angle of pi/6  and a speed of 2 m/s. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

Jul 2, 2017

$\text{distance} = 0.184$ $\text{m}$

#### Explanation:

We're asked to find the distance a projectile is from its starting point when it reaches its maximum height, given its initial velocity.

When the projectile is at its maximum height, its instantaneous $y$-velocity ${v}_{y}$ is zero. We can use the equation

${v}_{y} = {v}_{0 y} - g t$

to find the time $t$ when this occurs.

First, let's find the components of the initial velocity, for reference:

v_(0x) = v_0cosalpha_0 = (2color(white)(l)"m/s")cos(pi/6) = color(red)(1.73 color(red)("m/s"

v_(0y) = v_0sinalpha_0 = (2color(white)(l)"m/s")sin(pi/6) = color(green)(1.00 color(green)("m/s"

Plugging in our known values into the equation (${v}_{y} = 0$ and $g = 9.81$ ${\text{m/s}}^{2}$), we have

0 = color(green)(1.00 $\textcolor{g r e e n}{{\text{m/s") - (9.81color(white)(l)"m/s}}^{2}} t$

$t = 0.102$ $\text{s}$

The $x$- and $y$- positions of at this time are given by

$\Delta x = {v}_{0 x} t$

$\Delta y = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$

Plugging in known values, we have

Deltax = (color(red)(1.73)color(white)(l)color(red)("m/s"))(0.102color(white)(l)"s")

$= 0.177$ $\text{m}$

Deltay = (color(green)(1.00)color(white)(l)color(green)("m/s"))(0.102color(white)(l)"s") - 1/2(9.81color(white)(l)"m/s"^2)(0.102color(white)(l)"s")^2

$= 0.0510$ $\text{m}$

The distance from the starting point is thus

$r = \sqrt{{\left(0.177 \textcolor{w h i t e}{l} \text{m")^2 + (0.0510color(white)(l)"m}\right)}^{2}}$

= color(blue)(0.184 color(blue)("m"